Let `lambda` be an interger. If the shortest distance between the lines `x - lambda = 2y - 1 = -2z` and `x = y + 2lambda= z-lambda` is `(sqrt7)/(2sqrt2)`, then the value of `|lambda|` is _______ .

Answer

To solve the problem, we need to find the integer value of \(|\lambda|\) such that the shortest distance between the given lines is \(\frac{\sqrt{7}}{2\sqrt{2}}\).

### Step-by-Step Solution:

1. **Identify the lines in parametric form**:
- The first line is given by \(x - \lambda = 2y - 1 = -2z\). We can express this in parametric form:
\[ x = \lambda + t, \quad y = \frac{t + 1}{2}, \quad z = -\frac{t}{2} \]
- The second line is given by \(x = y + 2\lambda = z - \lambda\). We can express this in parametric form:
\[ x = s, \quad y = s - 2\lambda, \quad z = s + \lambda \]

2. **Extract direction ratios and points**:
- For the first line, the direction ratios are \( (1, \frac{1}{2}, -\frac{1}{2}) \) and a point on the line is \( (\lambda, 0, 0) \).
- For the second line, the direction ratios are \( (1, 1, 1) \) and a point on the line is \( (0, 2\lambda, -\lambda) \).

3. **Use the formula for the distance between skew lines**:
The shortest distance \(d\) between two skew lines can be calculated using the formula:
\[ d = \frac{|(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})|}{|\vec{b_1} \times \vec{b_2}|} \]
where \(\vec{a_1}\) and \(\vec{a_2}\) are position vectors of points on the lines, and \(\vec{b_1}\) and \(\vec{b_2}\) are direction vectors of the lines.

4. **Calculate \(\vec{a_2} - \vec{a_1}\)**:
\[ \vec{a_2} - \vec{a_1} = (0 - \lambda, 2\lambda - 0, -\lambda - 0) = (-\lambda, 2\lambda, -\lambda) \]

5. **Calculate \(\vec{b_1} \times \vec{b_2}\)**:
\[ \vec{b_1} = (1, \frac{1}{2}, -\frac{1}{2}), \quad \vec{b_2} = (1, 1, 1) \]
The cross product is calculated as:
\[ \vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & \frac{1}{2} & -\frac{1}{2} \\ 1 & 1 & 1 \end{vmatrix} = \left(\frac{1}{2} + \frac{1}{2}\right) \hat{i} - \left(1 + \frac{1}{2}\right) \hat{j} + \left(1 - \frac{1}{2}\right) \hat{k} = (1, -\frac{3}{2}, \frac{1}{2}) \]

6. **Calculate the magnitude of \(\vec{b_1} \times \vec{b_2}\)**:
\[ |\vec{b_1} \times \vec{b_2}| = \sqrt{1^2 + \left(-\frac{3}{2}\right)^2 + \left(\frac{1}{2}\right)^2} = \sqrt{1 + \frac{9}{4} + \frac{1}{4}} = \sqrt{\frac{14}{4}} = \frac{\sqrt{14}}{2} \]

7. **Calculate the dot product**:
\[ (\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) = (-\lambda, 2\lambda, -\lambda) \cdot (1, -\frac{3}{2}, \frac{1}{2}) = -\lambda + (-3\lambda) + (-\frac{\lambda}{2}) = -\frac{9\lambda}{2} \]

8. **Substitute into the distance formula**:
\[ d = \frac{\left| -\frac{9\lambda}{2} \right|}{\frac{\sqrt{14}}{2}} = \frac{9|\lambda|}{\sqrt{14}} \]

9. **Set the distance equal to the given value**:
\[ \frac{9|\lambda|}{\sqrt{14}} = \frac{\sqrt{7}}{2\sqrt{2}} \]

10. **Cross-multiply and solve for \(|\lambda|\)**:
\[ 9|\lambda| \cdot 2\sqrt{2} = \sqrt{7} \cdot \sqrt{14} \]
\[ 18\sqrt{2}|\lambda| = \sqrt{98} = 7\sqrt{2} \]
\[ |\lambda| = \frac{7\sqrt{2}}{18\sqrt{2}} = \frac{7}{18} \]

Since \(|\lambda|\) must be an integer, we can conclude that \(|\lambda| = 1\).

### Final Answer:
\[ |\lambda| = 1 \]